Question

在我的请求处理程序中，我想进行一些验证，并根据验证检查的结果，我将返回不同的响应（成功/错误）。因此，我为响应对象创建了一个抽象类，并为失败案例和成功案例创建了2个子类。代码看起来像这样，但它不编译，抱怨errorResponse和successResponse无法转换为AbstractResponse。

我是Java Generic和Spring MVC的新手，所以我不知道解决这个问题的简单方法。

@ResponseBody ResponseEntity<AbstractResponse> createUser(@RequestBody String requestBody) {
    if(!valid(requestBody) {
        ErrorResponse errResponse = new ErrorResponse();
        //populate with error information
        return new ResponseEntity<> (errResponse, HTTPStatus.BAD_REQUEST);
    }
    createUser();
    CreateUserSuccessResponse successResponse = new CreateUserSuccessResponse();
    // populate with more info
    return new ResponseEntity<> (successResponse, HTTPSatus.OK);
}

Answer 1

这里有两个问题：

必须更改您的返回类型以匹配两个响应子类ResponseEntity<? extends AbstractResponse>

当您实例化您的ResponseEntity时，您无法使用简化的＆lt;＆gt;语法必须指定要使用的响应类new ResponseEntity<ErrorResponse> (errResponse, HTTPStatus.BAD_REQUEST);

@ResponseBody ResponseEntity<? extends AbstractResponse> createUser(@RequestBody String requestBody) {
    if(!valid(requestBody) {
        ErrorResponse errResponse = new ErrorResponse();
        //populate with error information
        return new ResponseEntity<ErrorResponse> (errResponse, HTTPStatus.BAD_REQUEST);
    }
    createUser();
    CreateUserSuccessResponse successResponse = new CreateUserSuccessResponse();
    // populate with more info
    return new ResponseEntity<CreateUserSuccessResponse> (successResponse, HTTPSatus.OK);
}

Answer 2

另一种方法是使用错误处理程序

@ResponseBody ResponseEntity<CreateUserSuccessResponse> createUser(@RequestBody String requestBody) throws UserCreationException {
    if(!valid(requestBody) {
        throw new UserCreationException(/* ... */)
    }
    createUser();
    CreateUserSuccessResponse successResponse = new CreateUserSuccessResponse();
    // populate with more info
    return new ResponseEntity<CreateUserSuccessResponse> (successResponse, HTTPSatus.OK);
}

public static class UserCreationException extends Exception {
    // define error information here
}

@ExceptionHandler(UserCreationException.class)
@ResponseStatus(HttpStatus.BAD_REQUEST)
@ResponseBody
public ErrorResponse handle(UserCreationException e) {
    ErrorResponse errResponse = new ErrorResponse();
    //populate with error information from the exception
    return errResponse;
}

这种方法可以返回任何类型的对象，因此不再需要成功案例和错误案例（甚至案例）的抽象超类。

Spring MVC：如何在ResponseEntity体中返回不同的类型

2 个答案: